Bending of Light around a Star

[kmarinas86]

I know that the formula for the angle of bending of light around the star $\theta \;$ in radians is:

$\theta \; = \frac{4GM}{rc^2}$

This is equal to full bending of light, half of which occurs on the rear side of the star and the other half occurring on the foreground side.

So half of this is equal to:

$\frac{\theta \;}{2} = \frac{2GM}{rc^2}$

So that:

$dt = \frac{d\tau \;}{\sqrt{1-\frac{\theta\;}{2}}}$

$\sqrt{1-\frac{\theta \;}{2}} = \frac{d\tau\;}{dt}$

$1-\frac{\theta \;}{2} = \frac{d\tau\;^2}{dt^2}$

$1-\frac{d\tau \;^2}{dt^2} = \frac{\theta\;}{2}$

$2 \left( 1-\frac{d\tau \;^2}{dt^2} \right) = \theta\;$

If $\theta \;=2 radians$ then there would be infinite gravitational time dilation.

Did I get that right?

 

[pervect]

I'm not sure about your notation. Also, the formula is only valid for small angles.

My text gives the angle of deflection (converting to non-geometric units) as

$$\Theta = \frac{4GM}{c^2b}$$

where b is the impact parameter, related to the Schwarzschild radius of the 'turning point" $r_{tp}$ by

$$b = \frac{r_{tp}}{\sqrt{1-\frac{2GM}{r_{tp}c^2}}}$$

This is only a small-angle approximation, the results won't be right for $\Theta = 2\pi$.

A photon that gets deflected by a black hole will always be outside the photon sphere at r=3GM/c^2, so the maximum time dilation at closest approach for the actual orbit will always be finite - less than $\sqrt{3}$.

[add]I've attached a numeric solution for a photon starting out just outside the photon radius at r_min=3.01, spiraling outwards. The mass M of the black hole is 1 in geometric units.

The appropriate differential equation used to generate it which describes the exact orbit of a photon around a Schwarzschild black hole is , using geometric units in which G=c=1:

$$\left( \frac{1}{r^2} \frac{d r}{d \theta} \right) ^2 + \frac{1-2M/r}{r^2} = \frac{1}{b^2} = \frac{1-2M/r_{min}}{r_{min}^2}$$